\( X \) is a normally distributed random variable with mean 81 and standard deviation 14 . What is the probability that \( X \) is between 40 and 64 ? Write your answer as a decimal rounded to the nearest thousandth.

To find the probability that \( X \) is between 40 and 64 for a normally distributed variable with a mean of 81 and a standard deviation of 14, we can standardize our variable using the Z-score formula: \[ Z = \frac{(X - \mu)}{\sigma} \] 1. For \( X = 40 \): \[ Z = \frac{(40 - 81)}{14} \approx \frac{-41}{14} \approx -2.93 \] 2. For \( X = 64 \): \[ Z = \frac{(64 - 81)}{14} \approx \frac{-17}{14} \approx -1.21 \] Now, we look up these Z-scores in the standard normal distribution table or use a calculator: - The probability of \( Z < -2.93 \) is approximately 0.0016. - The probability of \( Z < -1.21 \) is approximately 0.1131. To find the probability of \( X \) being between 40 and 64, we subtract the two probabilities: \[ P(40 < X < 64) = P(Z < -1.21) - P(Z < -2.93) \approx 0.1131 - 0.0016 = 0.1115 \] So the probability that \( X \) is between 40 and 64 is approximately **0.112** when rounded to the nearest thousandth.